# For a connected semisimple Lie group $G$, $\mathrm{Inn}(G)$ is a subgroup of finite index in $\mathrm{Aut}(G)$.

For a connected semisimple Lie group $$G$$, $$\mathrm{Inn}(G)$$ is a subgroup of finite index in $$\mathrm{Aut}(G).$$

Can anyone give reference or proof for it. I know reference for compact but in general I don't know.

First of all, let $${\mathfrak g}$$ denote the Lie algebra of $$G$$. Since $$G$$ is connected, linearlization at $$e\in G$$ defines an embedding $$\phi: Aut(G)\to Aut({\mathfrak g})$$ (which need not be surjective, unless $$G$$ is simply connected, but this is not our problem). The group $$Aut(G)$$ is a (possibly disconnected) Lie group, the image of its Lie algebra under $$d\phi$$ consists of derivations of $${\mathfrak g}$$. The $$d\phi$$-image of the Lie algebra of $$Inn(G)$$ (i.e. the image of adjoint representation of the Lie algebra of $$G$$) consists of inner derivations of $${\mathfrak g}$$. Every derivation of a semisimple Lie algebra is inner. Thus, the identity component of $$Aut(G)$$ coincides with $$Inn(G)$$ (since they have the same Lie algebra). It remains to prove that the identify component of $$Aut({\mathfrak g})$$ is a finite index subgroup in $$Aut({\mathfrak g})$$. This is proven in Corollary 2 of the linked Murakami's paper "On the automorphisms of a real semisimple Lie algebra", Journal of the Mathematical Society of Japan Vol. 4, No. 2, 1952.
Theorem: Let $$G$$ be a reductive group. Then, $$\mathrm{Out}(G)$$ is finite if and only if $$Z(G)^\circ$$ has rank at most $$1$$.